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The upper bound estimate of the number of integer points on elliptic curves {y}^{2}={x}^{3}+{p}^{2r}x
Journal of Inequalities and Applications volume 2014, Article number: 104 (2014)
Abstract
Let p be a fixed prime and r be a fixed positive integer. Further let N({p}^{2r}) denote the number of pairs of integer points (x,\pm y) on the elliptic curve E:{y}^{2}={x}^{3}+{p}^{2r}x with y>0. Using some properties of Diophantine equations, we give a sharper upper bound estimate for N({p}^{2r}). That is, we prove that N({p}^{2r})\le 1, except with N({17}^{2(2s+1)})=2, where s is a nonnegative integer.
MSC:11G05, 11Y50.
1 Introduction
Let ℤ, ℕ be the sets of all integers and positive integers, respectively. Let p be a fixed prime and k be a fixed positive integer. Recently, the integer points (x,y) on the elliptic curve
have been investigated in many papers (see [1, 2] and [3]). In this paper we deal with the number of integer points on (1.1) for even k. Then (1.1) can be rewritten as
where r is a positive integer.
An integer point (x,y) on (1.2) is called trivial or nontrivial according to whether y=0 or not. Obviously, (1.2) has only the trivial integer point (x,y)=(0,0). Notice that if (x,y) is a nontrivial integer point on (1.2), then (x,y) is also. Therefore, (x,y) along with (x,y) are called by a pair of nontrivial integer points and denoted by (x,\pm y), where y>0. For any positive integer n, let
where
Using some properties of Diophantine equations, we give a sharper upper bound estimate for N({p}^{2r}), the number of pairs of nontrivial integer points (x,\pm y) on (1.2). That is, we shall prove the following results.
Theorem 1.1 All nontrivial integer points on (1.2) are given as follows.

(i)
p=u({2}^{m}), r=2s+1, (x,\pm y)=({p}^{2s}{v}^{2}({2}^{m}),\pm {p}^{3s}v({2}^{m})({v}^{2}({2}^{m})+1)), where m, s are nonnegative integers.

(ii)
p\equiv 1(mod\phantom{\rule{0.25em}{0ex}}8), r=2s+1, (x,\pm y)=({p}^{2s+1}{X}^{2},\pm {p}^{3s+2}XY), where s is a nonnegative integer, (X,Y) is a solution of the equation
{X}^{4}p{Y}^{2}=1,\phantom{\rule{1em}{0ex}}X,Y\in \mathbb{N}.(1.5)
Theorem 1.2 Let p be an odd prime, r be a positive integer. Then for any nonnegative integer s, we have N({p}^{2r})\le 1, except with N({17}^{2(2s+1)})=2. Moreover, if p\not\equiv 1(mod\phantom{\rule{0.25em}{0ex}}8), then N({p}^{2r})=0, except with N({3}^{2(2s+1)})=1.
2 Lemmas
Lemma 2.1 ([[4], Theorem 244])
Every solution (u,v) of the equation
can be expressed as (u,v)=(u(n),v(n)), where n is a positive integer.
Lemma 2.2 If p=u(n), then n={2}^{m}, where m is a nonnegative integer.
Proof Assume that n has an odd divisor d with d>1. Then we have either u(1)u(n) and 1<u(1)<u(n) or u(n/d)u(n) and 1<u(n/d)<u(n). Therefore, since p is a prime, it is impossible. Thus, we get n={2}^{m}. The lemma is proved. □
Any fixed positive integer a can be uniquely expressed as a=b{c}^{2}, where b, c are positive integers with b is square free. Then b is called the quadratfrei of a and denoted by Q(a).
Lemma 2.3 For any positive integer m, we have 3Q(v({2}^{m})).
Proof By (1.3) and (1.4), we get
and
Since (2/3)=1, where (2/3) is the Legendre symbol, we see from (2.3) that 3\nmid u({2}^{i}) for i\ge 1. Therefore, since u(1)=3, by (2.2), we obtain 3\parallel v({2}^{m}). It implies that 3Q(v({2}^{m})). The lemma is proved. □
Let D be a nonsquare positive integer. It is a well known fact that if the equation
has solutions (U,V), then it has a unique solution ({U}_{1},{V}_{1}) such that {U}_{1}+{V}_{1}\sqrt{D}\le U+V\sqrt{D}, where (U,V) through all solutions of (2.4). For any odd positive integer l, let
Then (U,V)=(U(l),V(l)) (l=1,3,\dots) are all solutions of (2.4).
Lemma 2.4 ([[5], Theorem 1])
The equation
has at most one solution (X,Y). Moreover, if the solution (X,Y) exists, then ({X}^{2},Y)=(U(l),V(l)), where l=Q({U}_{1}).
Lemma 2.5 ([[5], Theorem 3])
If 3Q({U}_{1}), then (2.5) has no solutions (X,Y).
Lemma 2.6 If p=u({2}^{m}), where m is a positive integer with m>1, then (1.5) has no solutions (X,Y).
Proof Since p=u({2}^{m}) with m>1, by (2.3), we have
We see from (2.6) that the equation
has solution (U,V) and its fundamental solution is ({U}_{1},{V}_{1})=(2v({2}^{m1}),1). Further, since m1\ge 1, by Lemma 2.3, we have 3Q(v({2}^{m1})). Hence, we get 3Q({U}_{1})=Q(2v({2}^{m1})). Therefore, by Lemma 2.5, the lemma is proved. □
Lemma 2.7 ([6])
The equation
has no solutions (X,Y,n).
Lemma 2.8 The equation
has only the solutions (p,X,Y,n)=(u({2}^{m}),{v}^{2}({2}^{m})+1,v({2}^{m}),1), where m is a nonnegative integer.
Proof Assume that (p,X,Y,n) is a solution of (2.9). If p=2, since gcd(X,Y)=1, then we have 2\nmid XY, gcd(X+{Y}^{2},X{Y}^{2})=2, X+{Y}^{2}={2}^{2n1}, X{Y}^{2}=2 and {Y}^{2}={2}^{2n2}1. But since {Y}^{2}+1 is not a square, it is impossible.
If p is an odd prime, then we have gcd(X+{Y}^{2},X{Y}^{2})=1, and by (2.9), we get X+{Y}^{2}={p}^{2n}, X{Y}^{2}=1,
and
By Lemma 2.7, we get from (2.11) that n=1 and
Further, applying Lemma 2.1 to (2.12) yields
Further, by Lemma 2.2, we see from the first equality of (2.13) that n={2}^{m}. Thus, by (2.10) and (2.13), the lemma is proved. □
3 Proof of Theorem 1.1
Assume that (x,\pm y) is a pair of nontrivial integer points on (1.2). Since y>0, we have x>0 and x can be expressed as
Substituting (3.1) into (1.2) yields
We first consider the case that r>t. By (3.2), we have
Since p\nmid z, we have p\nmid {z}^{2}+{p}^{2r2t} and gcd(z,{z}^{2}+{p}^{2k2r})=1. Hence by (3.3), we get
whence we obtain
Applying Lemma 2.8 to (3.5) yields
Therefore, by (3.1), (3.4), and (3.6), the integer points of type (i) are given.
We next consider the case that r=t. Then we have
Since p\nmid z, gcd(z,{z}^{2}+1)=1 and {z}^{2}+1 is not a square, we see from (3.7) that
By (3.8), we get
It implies that (X,Y)=(f,g) is a solution of (1.5). Therefore, by (3.1) and (3.8), we obtain the integer points of type (ii).
We finally consider the case that r<t. Then we have
Since p\nmid z({p}^{2t2r}{z}^{2}+1) and gcd(z,{p}^{2t2r}{z}^{2}+1)=1, we see from (3.10) that {p}^{2t2r}{z}^{2}+1 is a square, a contradiction.
To sum up, the theorem is proved.
4 Proof of Theorem 1.2
By (2.3), if p=u({2}^{m}) with m\ge 1, then p\equiv 1(mod\phantom{\rule{0.25em}{0ex}}8). Therefore, by Theorem 1.1, if p\not\equiv 1(mod\phantom{\rule{0.25em}{0ex}}8), then (1.2) has only the nontrivial integer point
It implies that the theorem is true for p\not\equiv 1(mod\phantom{\rule{0.25em}{0ex}}8).
For p\equiv 1(mod\phantom{\rule{0.25em}{0ex}}8), let {N}_{1} and {N}_{2} denote the number of pairs of nontrivial integer points of types (i) and (ii) in Theorem 1.1, respectively. Obviously, we have
and {N}_{1}\le 1. By Lemma 2.4, we get {N}_{2}\le 1. Hence, by (4.2), we have N({p}^{2r})\le 2 for p\equiv 1(mod\phantom{\rule{0.25em}{0ex}}8). Since u(2)=17 and (1.5) has the solution (X,Y)=(2,1) for p=17, by Theorem 1.1, we get
and N({17}^{2(2s+1)})=2. However, by Lemma 2.6, if p=u({2}^{m}) with m>1, then {N}_{2}=0. Therefore, by (4.2), if p\equiv 1(mod\phantom{\rule{0.25em}{0ex}}8), then N({p}^{2r})\le 1, except with N({17}^{2(2s+1)})=2. The theorem is proved.
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Acknowledgements
The authors would like to thank the referees for their very helpful and detailed comments, which have significantly improved the presentation of this paper. This work is supported by the P.E.D. (2013JK0573) and N.S.F. (11371291) of P.R. China.
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JZ obtained the theorems and completed the proof. XL corrected and improved the final version. Both authors read and approved the final manuscript.
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Zhang, J., Li, X. The upper bound estimate of the number of integer points on elliptic curves {y}^{2}={x}^{3}+{p}^{2r}x. J Inequal Appl 2014, 104 (2014). https://doi.org/10.1186/1029242X2014104
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DOI: https://doi.org/10.1186/1029242X2014104
Keywords
 elliptic curve
 integer point
 Diophantine equation